Careers360 Logo
NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

Edited By Vishal kumar | Updated on Apr 02, 2025 12:36 PM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 04 Apr' 2025 - 04 Apr' 2025

Electricity is the chapter that brings the flow of electric charge to life, helping you understand how circuits work and why resistance matters in everyday electronics! This chapter is one of the most interesting and scoring topics in the Class 12 Physics syllabus. Here, you'll find easy-to-understand NCERT solutions for questions 3.1 to 3.9, along with a few important additional questions from the chapter to give you extra practice. These solutions are prepared by Physics experts and are fully aligned with the latest CBSE syllabus to help you learn and revise better.

This Story also Contains
  1. NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity
  2. Additional Questions
  3. Class 12 Physics NCERT Solutions for Current Electricity: Chapter- Wise
  4. Current Electricity Class 12: Important Formulas
  5. NCERT Class 12 Physics Chapter 3: Important Topics
  6. Importance of Solutions of NCERT for Class 12 Chapter 3 Current Electricity in Board Exams:
  7. Key Features of Class 12 Physics Chapter 3 Exercise Solutions
NCERT solutions for Class 12 Physics Chapter 3 Current Electricity
NCERT solutions for Class 12 Physics Chapter 3 Current Electricity
LiveJEE Main 2025 April 3 Exam LIVE: NTA JEE shift-1 exam begins; paper analysis, difficulty level, updatesApr 3, 2025 | 9:00 AM IST

The JEE Mains 2025 April 3 shift 1 exam commenced at 9 am. The JEE Mains April 3 shift 1 exam 2025 will conclude at 12 noon. 

Read More

These NCERT solutions are a great way to build your confidence and boost your exam preparation. They not only help you with board exam questions but are also super useful if you’re preparing for JEE, NEET, or other entrance exams. So, make sure to go through each solution carefully and practice the additional questions to strengthen your understanding of Current Electricity!

Download the free PDF of Class 12 Physics Chapter 3 Exercise Solutions and get exam-ready for your CBSE board! These detailed solutions cover all the important questions, helping you strengthen your understanding of Current Electricity and score high in exams.

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

Download PDF

NCERT solutions for class 12 physics chapter 3 Current Electricity: Exercise Question

3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω , what is the maximum current that can be drawn from the battery?

Answer :

Given, the emf of battery, E = 12 V

Internal resistance of battery, r = 0.4 Ohm

Let I be the maximum current drawn from the battery.

We know, according to Ohm's law

E = Ir

I = E/r = 12/0.4 =30 A

Hence the maximum current drawn from the battery is 30 A.

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer:

Given, The emf of the battery, E = 10 V

The internal resistance of the battery, r = 3 Ohm

Current in the circuit, I = 0.5 A

Let R be the resistance of the resistor.

Therefore, according to Ohm's law:

E = IR' = I(R + r)

10 = 0.5(R + 3)

R = 1 Ohm

Also,

V = IR (Across the resistor)

= 0.5 x 17 = 8.5 V

Hence, the terminal voltage across the resistor = 8.5 V

3.3 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.7 x 10-4 oC-1

Answer:

Given,

temperature coefficient of filament,

α = 1.7 x 10-4 oC-1

T1=27C ; R1=100Ω

Let T2 be the temperature of element, R2=117 Ω

(Positive alpha means that the resistance increases with temperature. Hence we can deduce that T2 will be greater than T1 )

We know,

R2=R1[1+αΔT]

117=100[1+(1.70×104)(T227)]

T227=1171001.7×104T227=1000T2=1027C

Hence, the temperature of the element is 1027 °C.

3.4 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0×107m2 , and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Answer:

R=ρl/A , where ρ is the resistivity of the material

ρ=RA/l=5×6×107/15

ρ=2×107

Hence, the resistivity of the material of wire is ρ=2×107 m

3.5 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Answer:

Given,

T1=27.5C ; R1=2.1Ω

T2=100C ; R2=2.7Ω

We know,

R2=R1[1+αΔT]

2.7 = 2.1[1 + α (100 - 27.5) ]

α = (2.7 - 2.1) / 2.1(100 -27.5)

α = 0.0039 C1

Hence, the temperature coefficient of silver wire is 0.0039 C1

3.6 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70×104C1.

Answer:

For the given voltage, the two values of current will correspond to two different values of resistance which will correspond to two different temperature.

V = 230 V

I1=3.2A and I2=2.8A

Using Ohm's law:

R1=230/3.2=71.87Ω

and

R2=230/2.8=82.14Ω

T1=27C

Let T2 be the steady temperature of the heating element.

We know,

R2=R1[1+αΔT]

That is

230/2.8 = 230/3.2[1 + ( 1.70×104 ) ( T 2 - 27) ]

T 2 = (840.5 + 27) °C

Hence, steady temperature of the element is 867.5 °C.

3.7 Determine the current in each branch of the network shown in Fig. 3.30:

image

Answer:

The current in the circuit is distributed like

image 1

where I1, I2, and I3 are the different currents through shown branches.

Now, applying KVL in Loop

10I10I25(I2+I3)10=0

Also, we have I=I1+I2

So putting it in KVL equation

10(I1+I2)10I25(I2+I3)10=0

1010I110I25I210I210I3=0

1010I125I210I3=0 .................................(1)

Now let's apply KVL in the loop involving I1 I2 AND I3

5I210I15I3=0 .................................(2)

now, the third equation of KVL

5I35(I1I3)+10(I2+I3)=0

5I1+10I2+20I3=0 ..............................(3)

Now we have 3 equation and 3 variable, on solving we get

I1=417A

I2=617A

I3=217A

Now the total current

I=I1+I2=417+617=1017

3.8 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer:

Emf of battery, E = 8 V

Internal resistance of battery, r = 0.5 Ω

Supply Voltage, V = 120 V

The resistance of the resistor, R = 15.5 Ω

Let V' be the effective voltage in the circuit.

Now, V' = V - E

V' = 120 - 8 = 112 V

Now, current flowing in the circuit is:

I = V' / (R + r)

I=11215.5+0.5=7A

Now, using Ohm ’s Law:

Voltage across resistor R is v = IR

V = 7 x 15.5 = 108.5 V

Now, the voltage supplied, V = Terminal voltage of battery + V

Terminal voltage of battery = 120 -108.5 = 11.5 V

The purpose of having a series resistor is to limit the current drawn from the supply.

3.9 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5×1028m3 . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0×106m2 and it is carrying a current of 3.0 A.

Answer:

We know,

I=neAvd

vd :drift Velocity = length of wire(l) / time taken to cover

I=neAlt

by substituting the given values

t = 2.7 x 104 s

Therefore, the time required by an electron to drift from one end of a wire to its other end is 2.7×104 s.

Additional Questions

1. The earth’s surface has a negative surface charge density of 10 -9 C m -2 . The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A cover the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10 6 m.)

Answer:

Given, the surface charge density of earth

ρ = 109Cm2

Current over the entire globe = 1800 A

Radius of earth, r = 6.37 x 106 m

The surface area of earth A = 4πr2

= 4π(6.37×106)2 = 5.09×1014m2

Now, charge on the earth surface,

q=ρ×A

Therefore,

q=1005.09×105C

Let the time taken to neutralize earth surface be t

Current I = q / t

t = 282.78 s.

Therefore, time take to neutralize the Earth's surface is 282.78 s

2. Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

Answer:

Given,

There are 6 secondary cells.

Emf of each cell, E = 2 V (In series)

The internal resistance of each cell, r = 0.015 Ω (In series)

And the resistance of the resistor, R = 8.5 Ω

Let I be the current drawn in the circuit.

I=nER+nr

I=6(2)8.5+6(0.015)=128.59

I = 1.4 A

Hence current drawn from the supply is 1.4 A

Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V

3. A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:

Given,

Emf , E = 1.9 V

Internal resistance, r =380 Ω

The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A

The motor requires a large value of current to start and hence this cell cannot be used for a motor of a car.

4. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. ( ρAl=2.63×108Ωm . ρCul=1.73×108Ωm relative density of Al = 2.7, of Cu = 8.9.)

Answer:

We know,

R = Ρ l / A

The wires have the same resistance and also are of the same length.

Hence,

ρAlAAl=ρCuACu

AAlACu=ρAlρCu=2.631.73

Now, mass = Density x Volume = Density x Area x length

Taking the ratio of their masses for the same length

mAlmCu=dALAAldCuACu=2.7×2.638.9×1.73<1

Hence, mAl<mCu

Therefore, for the same resistance and length, the aluminium wire is lighter.

Since aluminium wire is lighter, it is used as power cables.

5. What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current

A

Voltage

V

Current

A

Voltage

V

0.2

3.94

3.0

59.2

0.4

7.87

4.0

78.8

0.6

11.8

5.0

98.6

0.8

15.7

6.0

118.5

1.0

19.7

7.0

138.2

2.0

39.4

8.0

158.0

NEET/JEE Coaching Scholarship

Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Answer:

The ratio of Voltage to current for the various values comes out to be nearly constant which is around 19.7.

Hence the resistor made of alloy manganin follows Ohm's law.

Answer the following questions:

6. A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?

Answer:

The current flowing through the conductor is constant for a steady current flow.

Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.

Answer the following questions:

7. Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.

Answer:

No. Ohm’s law is not universally applicable for all conducting elements.

A semiconductor diode is such an example.

Answer the following questions

8. A low voltage supply from which one needs high currents must have very low internal resistance. Why?

Answer:

Ohm's law states that: V = I x R

Hence for a low voltage V, resistance R must be very low for a high value of current.

Answer the following questions:

9. A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer:

A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.

10. Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of ( 1022/1023 ).

Answer:

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 1022 .

11. (a)

(i) Given n resistors each of resistance R , how will you combine them to get the (i) maximum effective resistance?

Answer:

To get maximum effective resistance, combine them in series. The effective resistance will be nR.

11. (b)

(ii) Given n resistors each of resistance R, how will you combine them to get the (ii) minimum effective resistance?

Answer:

To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.

11.(c)

(iii) What is the ratio of the maximum to minimum resistance?

Answer:

The ratio is nR/(R/n) = n2

12.(a)

(i) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them
to get an equivalent resistance of (i) 113Ω

Answer:

We have, equivalent resistance = 11/3

Let's break this algebraically so that we can represent it in terms of 1, 2 and 3

113=9+23=3+23=3+121+2

this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :

12. (b)

(ii) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine the to get an equivalent resistance of (ii) 115Ω

Answer:

Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it

Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1

R = 11/5 Ω

12.(c)

(iii) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (iii) 6 Ω

Answer:

1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.

12.(d)

(iv) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them
to get an equivalent resistance of (iv) 611Ω

Answer:

Connect all three resistors in parallel.

Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)

R = 6/11 Ω

13.

(a) Determine the equivalent resistance of networks shown in
Fig. 3.31.

Answer:

It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops,

so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is

Equivalent Rloop=242+4=86=43

now we have 4 such loops in series so,

Total Equivalent Rloop=43+43+43+43=163

Hence equivalent resistance of the circuit is 16/3 ohm.

13.

(b) Determine the equivalent resistance of networks shown in Fig. 3.31.

Answer:

It can be seen that all 5 resistors are in series, so

Equivalent Resistance = R + R + R + R + R = 5R

Hence equivalent resistance is 5R.

14. Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

Answer:

First, let us find the equivalent of the infinite network,

let equivalent resistance = R'

Here from the figure, We can consider the box as a resistance of R'

Now, we can write,

equivalent resistance = R'

' =[( R')Parallel with (1)] + 1 + 1

R1R+1+2=R

R+2R+2=R2+R

R22R2=0

R=1+3,or13

Since resistance can never be negative we accept

R=1+3

, We have calculated the equivalent resistance of infinite network,

Now

Total equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network

= 0.5+1+1.73

=3.23 ohm

V=IR

I=VR=123.23=3.72A

Hence current drawn from the 12V battery is 3.72 Ampere.

15. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value of ϵ

Answer:

Given

Maintained constant emf of standard cell = 1.02V, balanced point of this cell = 67.3cm

Now when the standard cell is replaced by another cell with emf = ε, balanced point for this cell = 82.3cm

Now as we know the relation

εl=EL

ε=ELl=1.0267.382.3=1.247V

Hence, emf of another cell is 1.247V.

16. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(b) What purpose does the high resistance of 600 k Ω have?

Answer:

If a sufficiently high current passes through galvanometer then it can get damaged. So we limit the current by adding a high resistance of 600 k Ω.

17. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(c) Is the balance point affected by this high resistance?

Answer:

No, the Balance point is not affected by high resistance. High resistance limits the current to galvanometer wire. The balance point is obtained by moving the joe key on the potentiometer wire and current through potentiometer wire is constant. The balance point is the point when the current through galvanometer becomes zero. The only duty of high resistance is to supply limited constant current to potentiometer wire.

18. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

Answer:

No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when emf of the driving point is less than the other cell, there won't be any balance point in the wire.

19. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Answer:

No, the circuit would not work properly for very low order of Voltage because the balance points would be near point A and there will be more percentage error in measuring it. If we add series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.

20.

Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer:

Given,

the balance point of cell in open circuit = l1=76.3cm

value of external resistance added = R=9.5Ω

new balance point = l2=64.8cm

let the internal resistance of the cell be r.

Now as we know, in a potentiometer,

r=l1l2l2R

r=76.364.864.89.5=1.68Ω

Hence, the internal resistance of the cell will be 1.68 Ω.

Chapter 3 of Class 12 Physics NCERT Solutions is crucial for students preparing for board exams, JEE, and NEET. This chapter introduces fundamental concepts like electric current and resistance, which are not only important for scoring well in board exams but also essential for competitive exams like JEE and NEET. The solutions help build a strong foundation for these topics, and understanding them is key for advancing in physics and electrical engineering. So, mastering this chapter is vital for both exam success and future studies!

Class 12 Physics NCERT Solutions for Current Electricity: Chapter- Wise

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Current Electricity Class 12: Important Formulas

  • Current density

di=JdAcosθ=JdA

Where: is the angle between the normal to the area and direction of current

Relation between current density and electric field-

J=σE=Eρ

where = conductivity and = resistivity or specific resistance of the substance

  • Ohm’s Law

VIV=IR

  • Resistivity

ρ=mne2τ

Where:

m is the mass, n is the number of electrons per unit volume, e is the charge of the electron and is the relaxation time

  • Grouping of Resistance

Series Grouping of resistance

Req=R1+R2++Rn

Parallel Grouping of Resistance

1Req=1R1+1R2+..+1Rn

  • Heat developed in a resistor

The power developed = energy  time =i2R=iR=V2R

  • Kirchoff's Current Law(KCL)

1694165013898

i=0
i1+i3=i2+i4

  • Kirchoff's Voltage Law(KVL)

1694165016529

V=0i1R1+i2R2E1i3R3+E2+E3i4R4=0

  • Wheatstone's Bridge

1694165018817

PQ=RSVB=VD( Balanced condition )

NCERT Class 12 Physics Chapter 3: Important Topics

Current Electricity (Chapter 3) in Class 12 Physics is important for both board exams and competitive exams like JEE Main and NEET. The topics covered in this chapter are essential, and NCERT Solutions for this chapter will help you understand them thoroughly.

Here’s a breakdown of the key topics in Chapter 3 - Current Electricity:

1. Electric Current

This part explains the concept of electric current, drift velocity, mobility, and other related concepts. All the formulas and concepts here are crucial for both the CBSE board exams and competitive exams like JEE Main and NEET.

2. Ohm's Law

Ohm's Law shows the relationship between voltage and current at a given temperature. It also covers ohmic and non-ohmic conductors, which follow or don't follow the law. This section introduces resistance, and questions based on Ohm’s Law are included in the NCERT Solutions for Class 12 Physics Chapter 3.

3. Resistance and Resistivity

You’ll learn about resistance and resistivity and how they change with temperature. The color code of resistance is also explained here. Make sure you understand these concepts, as there are practice questions based on them in the NCERT solutions.

4. Cells and Their Combinations

This section explains the concepts of cells and how they work in series and parallel combinations. You’ll definitely encounter questions on this in your exams.

5. Kirchhoff's Laws

Here, you'll study Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL), which are essential for solving electrical circuits. These laws are important for competitive exams like JEE Main and NEET.

6. Wheatstone's Bridge, Meter Bridge, and Potentiometer

You’ll learn about the Wheatstone Bridge, the Meter Bridge and how they work, including their balanced conditions. The potentiometer is also introduced here, along with its applications, like comparing EMFs of cells and finding internal resistance. These topics have high weightage in exams, and you can expect questions on them in NEET and JEE Main. Practice questions from these topics using the NCERT Solutions to get better at them.

Importance of Solutions of NCERT for Class 12 Chapter 3 Current Electricity in Board Exams:

When it comes to CBSE board exams and competitive exams like NEET and JEE Main, the solutions for NCERT Class 12 Physics Chapter 3 are super important. 7-10% of the questions in NEET and JEE Main are based on this chapter. In the 2019 CBSE board exam, there was even a 6-mark question on Current Electricity. So, using these NCERT solutions for Chapter 3 will help you score better in both your board exams and competitive exams.

Benefits of NCERT Class 12 Physics Chapter 3 Solutions

  • These NCERT Solutions for Class 12 Physics Chapter 3 are perfect for practising questions for your annual exams.
  • You can solve the exercises on your own and then check your answers with the solutions provided.
  • These solutions will give you clear and accurate answers, helping you score better marks in the exams.
  • If you get stuck on any question while solving the exercises in Chapter 3, you can always refer to these solutions to understand how to approach the question and find the correct answer.

Key Features of Class 12 Physics Chapter 3 Exercise Solutions

  • Complete Coverage: These solutions cover all topics and questions in Chapter 3 - Current Electricity, making sure you understand everything clearly.

  • Easy-to-understand Explanations: Each solution breaks down concepts step-by-step, so even tough topics are easier to grasp.

  • Simple and Clear: The solutions are written in simple, clear language, which helps you understand the concepts without confusion.

  • Practice Questions: The exercises include practice questions that let you test yourself and check your understanding.

  • Perfect for Exam Prep: These solutions help you prepare well for your board exams, JEE, NEET, and any other competitive exams.

  • Foundation for Future Learning: The concepts in this chapter are essential for understanding advanced topics in physics and electrical engineering.

  • Links to Later Topics: The chapter connects to future chapters, showing how important it is as a foundation for your studies.

  • Free Access: You can get these solutions for free, making them easily available for everyone.

These features make the NCERT Class 12 Physics Chapter 3 Solutions a super helpful resource to help you ace your exams and build a strong understanding of physics!

Also, check the NCERT Books and NCERT Syllabus here:

Subject wise solutions

Also, check

Frequently Asked Questions (FAQs)

1. How important is the NCERT Solutions for Class 12 Physics chapter 3 for CBSE board exam?

In CBSE board exam, around 8 to 10% questions can be expected from the chapter Current Electricity. Certain papers of CBSE ask around 15% questions from NCERT chapter 3 Current Electricity. To score well in the exam follow NCERT syllabus and the exercise given in the NCERT Book. To practice more problems can refer to NCERT exemplar.

2. What are the important topics of Current Electricity?

If we analyse the previous year NEET, JEE Main and board papers we can see that the topics like meter bridge, potentiometer and problems using kcl and kvl and combinations of resistance and cells are frequently asked.

3. What is the weightage of current electricity in JEE main?

In JEE main, 2 to 3 questions can be expected from the chapter on Current Electricity. The topics like Potentiometer, meter bridge, KVL and KCL are important.

4. Whether NCERT is enough to cover Current Electricity for NEET?

The concepts in the current electricity chapter of NCERT Class 12 Physics are integral parts of NEET exam. Along with NCERT, practicing with previous year papers and mock tests would be enough.

5. What is the weightage of current electricity for NEET exam?

Current electricity carries 8% of weightage on an average for NEET exam.

6. How important is the NCERT Solutions for Class 12 Physics chapter 3 for higher studies in the field of engineering?

Current electricity is the basics of electrical and electronics engineering-related branches. In these branches, analysis and designs of the circuit are important and the basic laws studied in the Class 12 Physics Chapter 3 NCERT solutions help for the same.

7. What are the main topics covered in Current Electricity Class 12 NCERT chapter?

The main topics are-

  • Concept of current
  • Drift velocity and mobility
  • Ohm's Law
  • Resistance and resistivity
  • Combination of resistors
  • Cells and combination of cells
  • Kirchhoff Law
  • Whetstones bridge, meter bridge and potentiometers
8. How can I score well in Current Electricity for my exams?

Focus on understanding the key concepts, practising numerical problems, and solving previous year questions. Refer to NCERT solutions for detailed explanations and practice regularly.

9. What are Kirchhoff’s laws, and why are they important?

Kirchhoff’s laws — Current Law (KCL) and Voltage Law (KVL) — help in analyzing complex electrical circuits. These laws are crucial for solving circuit problems, especially in JEE and NEET exams.

10. What is the significance of the Wheatstone Bridge and the Potentiometer?

The Wheatstone Bridge and Potentiometer are important for measuring resistance and EMF in electrical circuits. They have a high weightage in competitive exams like JEE Main and NEET.

Articles

Explore Top Universities Across Globe

University of Essex, Colchester
 Wivenhoe Park Colchester CO4 3SQ
University College London, London
 Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
 Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
 Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
 University Park, Nottingham NG7 2RD

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top